Find the slope of the line through (-6,-3) and (0,3)

Answers

Answer 1

Answer: slope is rise over run
rise (delta y) run (delta x)
delta= change in

rise 3-(-3) = 6
run 0-(-6) = 6

6/6 = 1,
slope is 1

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The midpoint of a line segment is (−3,2). One endpoint of the line segment is (4,−7).What are the coordinates of the other endpoint of the line segment?

Answers

The coordinates of the other endpoint of the line segment are $Y(x,y) = (-10, 11)$ .

Vectorially speaking, the midpoint of a line segment is determined by the following expression:

$M(x,y) = (1)/(2)\cdot X(x,y) + (1)/(2)\cdot Y(x,y)$ (1)

Where:

$M(x,y)$ - Midpoint.
$X(x,y)$ , $Y(x,y)$ - Endpoints of the line segment.

If we know that $M(x,y) = (-3, 2)$ and $X(x,y) = (4, -7)$ , then the coordinates of the other endpoint is:

$(1)/(2)\cdot Y(x,y) = M(x,y) -(1)/(2)\cdot X(x,y)$

$Y(x,y) = 2\cdot M(x,y) -X(x,y)$

$Y(x,y) = 2\cdot (-3,2)-(4,-7)$

$Y(x,y) = (-6, 4) - (4, -7)$

$Y(x,y) = (-10, 11)$

The coordinates of the other endpoint of the line segment are $Y(x,y) = (-10, 11)$ .

We kindly invite to check this question on midpoints: brainly.com/question/17506315

I need to find an equation that hass an infinite number of solutions

Answers

Answer

2(7x + 3) - x

= 13x + 6

Explanation

2(7x + 3) - x

= 14x + 6 - x

= 14x - x + 6

= 13x + 6

Hope this Helps!!!

Gloria earned a total of $810 over the summer. She earned $162 babysitting and the restfrom mowing lawns. Which bar model and solution represent x, the amount of money Gloria
earned mowing lawns?

Answers

Let x represent the amount of money Gloria earned mowing lawns.

We have been given that Gloria earned a total of $810 over the summer. She earned $162 babysitting and the rest from mowing lawns.

The total amount earned by Gloria would be amount earned from baby sitting and lawn mowing that is $x+162$ .

Now we will equate total earnings of Gloria by 810 as:

$x+162=810$

$x+162-162=810-162$

$x=648$

Therefore, Gloria earned $648 from mowing lawns.

Answer:

Gloria earned $648 from mowing lawns

Step-by-step explanation:

will give 5 stars and thanks for correct answer Richard starts high school every day at 7:45 A.M.. How many seconds is Richard in school each day of school dismissed at 2:15 P.M.

Answers

Richard spends 21,600 seconds at school each day.

(1 point) For the given position vectors r(t)r(t), compute the (tangent) velocity vector r′(t)r′(t) for the given value of tt . A) Let r(t)=(cos4t,sin4t)Let r(t)=(cos⁡4t,sin⁡4t). Then r′(π4)r′(π4)= ( , )? B) Let r(t)=(t2,t3)Let r(t)=(t2,t3). Then r′(5)r′(5)= ( , )? C) Let r(t)=e4ti+e−5tj+tkLet r(t)=e4ti+e−5tj+tk. Then r′(−5)r′(−5)= i+i+ j+j+ kk ?

Answers

Answer:

(a)

$r'(\frac \pi 4) =(0.-4)$

(b)

r'(5)= (10,75)

(c)

$r'(-5) =4 e^(-20)\hat i-5e^(25)\hat j+\hat k$

Step-by-step explanation:

(a)

Give that,the position vector is

r(t) = (cos 4t, sin 4t)

Differentiating with respect to t

r'(t) = (-4sin 4t, 4 cos 4t) [ $(d)/(dt) cos mt = -m \ sin \ mt$ and $(d)/(dt) sin mt = m \ cos \ mt$ ]

To find the $r'(\frac\pi 4)$ , we put $t=\frac \pi4$

$r'(\frac\pi 4) = (-4sin (4.\frac \pi 4), 4 cos (4.\frac \pi 4))$

=(0, -4)

(b)

Give that,the position vector is

r(t) = (t²,t³)

Differentiating with respect to t

r'(t) = (2t, 3t²)

To find r'(5) , we put t=5

r'(5) = (2.5,3.5²)

= (10,75)

(c)

Given position vector is

$r(t) = e^(4t)\hat i+e^(-5t)\hat j+t\hat k$

Differentiating with respect to t

$r'(t) =4 e^(4t)\hat i+(-5)e^(-5t)\hat j+\hat k$

$\Rightarrow r'(t) =4 e^(4t)\hat i-5e^(-5t)\hat j+\hat k$

To find r'(-5) , we put t= - 5 in the above equation

$r'(-5) =4 e^(4.(-5))\hat i-5e^(-5.(-5))\hat j+\hat k$

$\Rightarrow r'(-5) =4 e^(-20)\hat i-5e^(25)\hat j+\hat k$

For the given position vectors r(t)r(t), compute the (tangent) velocity vector r′(t)r′(t) for the given value of tt are:

$A) r' (\pi /4) = (0, -4) \nB) r'(5) = (10, 75)\nC) r'(-5) = (4e^(-20), -5e^(25), 1)$

To compute the velocity vector, we need to find the derivative of the position vector with respect to time (t). This will give us the tangent velocity vector.

A) Let r(t) = (cos⁡4t, sin⁡4t).

To find r'(t), we take the derivative of each component with respect to t:

r'(t) = (d/dt (cos⁡4t), d/dt (sin⁡4t))

r'(t) = (-4sin⁡4t, 4cos⁡4t)

To find r'(π/4), we substitute t = π/4 into r'(t):

r'(π/4) = (-4sin⁡(4(π/4)), 4cos⁡(4(π/4)))

r'(π/4) = (-4sin⁡π, 4cos⁡π)

r'(π/4) = (0, -4)

B) $Let \ r(t) = (t^2, t^3).$